Difference between revisions of "1980 AHSME Problems/Problem 28"

Latest revision as of 19:04, 30 October 2021

Problem

The polynomial $x^{2n}+1+(x+1)^{2n}$ is not divisible by $x^2+x+1$ if $n$ equals

$\text{(A)} \ 17 \qquad \text{(B)} \ 20 \qquad \text{(C)} \ 21 \qquad \text{(D)} \ 64 \qquad \text{(E)} \ 65$

Solution 1

Let $h(x)=x^2+x+1$ .

Then we have $(x+1)^2n = (x^2+2x+1)^n = (h(x)+x)^n = g(x) \cdot h(x) + x^n,$ where $g(x)$ is $h^{n-1}(x) + nh^{n-2}(x) \cdot x + ... + x^{n-1}$ (after expanding $(h(x)+x)^n$ according to the Binomial Theorem).

Notice that $x^2n = x^2n+x^{2n-1}+x^{2n-2}+...x -x^{2n-1}-x^{2n-2}-x^{2n-3} +...$

$x^n = x^n+x^{n-1}+x^{n-2} -x^{n-1}-x^{n-2}-x^{n-3} +....$

Therefore, the left term from $x^2n$ is $x^{(2n-3u)}$

          the left term from  is ,

If divisible by h(x), we need 2n-3u=1 and n-3v=2 or

                             2n-3u=2 and n-3v=1

The solution will be n=1 or 2 mod(3). Therefore n=21 is impossible

~~Wei

Solution 2

Notice that the roots of $w^2+w+1=0$ are also the third roots of unity (excluding $w=1$ ). This is fairly easy to prove: multiply both sides by $w-1$ and we get $(w-1)(w^2+w+1) = w^3 - 1 = 0.$ These roots are $w = e^{i \pi /3}$ and $w = e^{2i \pi /3}$ .

Now we have $\begin{align*} w^{2n} + 1 + (w+1)^{2n} &= w^{2n} + 1 + (-w^2)^{2n} \\ &= w^{4n} + w^{2n} + 1\\ &= 0. \end{align*}$ Plug in the roots of $w^2+w+1=0$ . Note that $e^{2i \pi /3} + 1 = -\frac{1}{2} + \frac{\sqrt{3}}{2}i + 1 = \frac{1}{2} + \frac{\sqrt{3}}{2}i = e^{i \pi /3}.$ However, this will not work if $n=3m$ , so $n$ cannot be equal to $21$ . Hence our answer is $\textrm{(C)}$ .

Solution 3

We start by noting that $x + 1 \equiv -x^2 \mod (x^2+x+1).$ Let $n = 3k+r$ , where $r \in \{ 0,1,2 \}$ .

Thus we have $x^{4n} + x^{2n} + 1 \equiv x^{4r} + x^{2r} + 1 \mod (x^3 -1).$

When $r = 0$ , $x^{4n} + x^{2n} + 1 \equiv 3 \mod (x^3 -1).$ When $r = 1$ , $x^{4n} + x^{2n} + 1 \equiv x^2 + x + 1 \mod (x^3 -1),$ which will be divisible by $x^2+x+1$ .

When $r = 2$ , $x^{4n} + x^{2n} + 1 \equiv x^2 + x + 1 \mod (x^3 -1),$ which will also be divisible by $x^2+x+1$ .

Thus $r \ne 0$ , so $n$ cannot be divisible by $3$ , and the answer is $\textrm{(C)}$ .

@@ Line 9: / Line 9: @@
 \text{(E)} \ 65  </math>
-== Solution ==
+== Solution 1==
-<math>\fbox{}</math>
+Let <math>h(x)=x^2+x+1</math>.
+Then we have
+<cmath>(x+1)^2n = (x^2+2x+1)^n = (h(x)+x)^n = g(x) \cdot h(x) + x^n,</cmath>
+where <math>g(x)</math> is <math>h^{n-1}(x) + nh^{n-2}(x) \cdot x + ... + x^{n-1}</math> (after expanding <math>(h(x)+x)^n</math> according to the Binomial Theorem).
+Notice that
+<cmath>x^2n = x^2n+x^{2n-1}+x^{2n-2}+...x
+           -x^{2n-1}-x^{2n-2}-x^{2n-3}
+         +...</cmath>
+<math>x^n = x^n+x^{n-1}+x^{n-2}
+         -x^{n-1}-x^{n-2}-x^{n-3}
+  +....</math>
+Therefore, the left term from <math>x^2n</math> is <math>x^{(2n-3u)}</math>
+           the left term from <math>x^n</math> is <math>x^{(n-3v)}</math>,
+If divisible by h(x), we need 2n-3u=1 and n-3v=2  or
+n-3u=2 and n-3v=1
+The solution will be n=1 or 2 mod(3). Therefore n=21 is impossible
+~~Wei
+==Solution 2==
+Notice that the roots of <math>w^2+w+1=0</math> are also the third roots of unity (excluding <math>w=1</math>). This is fairly easy to prove: multiply both sides by <math>w-1</math> and we get <cmath>(w-1)(w^2+w+1) = w^3 - 1 = 0.</cmath> These roots are <math>w = e^{i \pi /3}</math> and <math>w = e^{2i \pi /3}</math>.
+Now we have
+<cmath>\begin{align*}
+w^{2n} + 1 + (w+1)^{2n} &= w^{2n} + 1 + (-w^2)^{2n} \\
+&= w^{4n} + w^{2n} + 1\\
+&= 0.
+\end{align*}</cmath>
+Plug in the roots of <math>w^2+w+1=0</math>. Note that
+<cmath>e^{2i \pi /3} + 1 = -\frac{1}{2} + \frac{\sqrt{3}}{2}i + 1 = \frac{1}{2} + \frac{\sqrt{3}}{2}i = e^{i \pi /3}.</cmath>
+However, this will not work if <math>n=3m</math>, so <math>n</math> cannot be equal to <math>21</math>. Hence our answer is <math>\textrm{(C)}</math>.
+==Solution 3==
+We start by noting that <cmath>x + 1 \equiv -x^2 \mod (x^2+x+1).</cmath>
+Let <math>n = 3k+r</math>, where <math>r \in \{ 0,1,2 \}</math>.
+Thus we have <cmath>x^{4n} + x^{2n} + 1 \equiv x^{4r} + x^{2r} + 1 \mod (x^3 -1).</cmath>
+When <math>r = 0</math>, <cmath>x^{4n} + x^{2n} + 1 \equiv 3 \mod (x^3 -1).</cmath>
+When <math>r = 1</math>, <cmath>x^{4n} + x^{2n} + 1 \equiv x^2 + x + 1 \mod (x^3 -1),</cmath> which will be divisible by <math>x^2+x+1</math>.
+When <math>r = 2</math>, <cmath>x^{4n} + x^{2n} + 1 \equiv x^2 + x + 1 \mod (x^3 -1),</cmath> which will also be divisible by <math>x^2+x+1</math>.
+Thus <math>r \ne 0</math>, so <math>n</math> cannot be divisible by <math>3</math>, and the answer is <math>\textrm{(C)}</math>.
 == See also ==

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
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